Question

In the reaction, $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ the concentration of $H_2$, $I_2$ and $HI$ at equilibrium are 8.0, 3.0 and 28 moles per litre respectively. What will be the equilibrium constant ?

• A. 30.61
• B. 32.66
• C. 29.4
• D. 20.9

Answer 1

Answer: (B)

32.66

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
Applying law of mass action,
$K_c=$ $\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
Given : $\left[H_2\right]=8.0mol{L}^{-1}$, $\left[I_2\right]=3.0mol{L}^{-1}$ ,$\left[HI\right]=28.0mol{L}^{-I}$
So, $K_c=\frac{(28.0{)}^2}{\left(8.0\right)\times \left(3.0\right)}=32.66$

Hence, option B is correct.